Calculating Fluid Eflux: The Impact of a Cylindrical Tank on a Platform

AI Thread Summary
The discussion focuses on calculating the speed of water eflux from a cylindrical tank and the time it takes for the tank to empty. The initial height of water is 3m, and the tank is 6m high, with a plug area of 3 cm² removed. For the speed of the water striking the ground, the participants reference the equation H=0.5gt² to determine the time and subsequently the velocity. The second part addresses the changing velocity of eflux, suggesting the use of the equation D(vol)/Dt = A*(2gh)^(0.5) to find the rate of change of volume over time. The conversation emphasizes the need for accurate calculations to determine both the speed and the time for the tank to empty.
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A cylindrical tank of diameter 90 cm rests on top of a platform 6m
high. Initially the tank is filled with water to a depth of 3m. A plug
whose are is 3 cm^2 is removed from the orifice on the side of the
tank at the bottom.

1) at what speed will water strike the ground?
11) how long will it take for the tank to be empty?
 
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What are your thoughts on this? Anything you got so far?
 
Yeah, for part one:
H=0.5gt^2
t^2 = 12/9.8
t= (12/9.8)^.5

v=u +at
v= 0 + (12*9.8)^0.5
this we could acheive

But for part 2:
Velocity of eflux changes with time. thus
D(vol)/Dt = A*(2gh)^.5 = ((Pi*d^2)/4)*dh/dt
Is this correct?
 
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